The 'factorial function' is defined as...
$\displaystyle \displaystyle z! = \int_{0}^{\infty} t^{z}\ e^{-t}\ dt = \Gamma(z+1)= z\ \Gamma(z)$ (1)
... and Your request [if I undestood correctly...] is to find the Mc Laurin series of $\displaystyle \ln z!$. Very well!... let's start from the well known 'inifinite product'...
$\displaystyle \displaystyle \frac{1}{\Gamma (z)} = z\ e^{\gamma\ z}\ \prod_{n=1}^{\infty} (1+\frac{z}{n})\ e^{-\frac{z}{n}}$ (2)
... where...
$\displaystyle \displaystyle \gamma = \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{k} - \ln n$ (3)
... is the so called 'Euler's constant'. From (1) and (2) we derive...
$\displaystyle \displaystyle \frac{1}{z!} = e^{\gamma\ z}\ \prod_{n=1}^{\infty} (1+\frac{z}{n})\ e^{-\frac{z}{n}}$ (4)
... and from (4)...
$\displaystyle \displaystyle \lambda(z)= \ln z! = -\gamma\ z - \sum_{n=1}^{\infty} \{\ln (1+\frac{z}{n}) - \frac{z}{n} \}$ (5)
Deriving both terms of (5) we obtain...
$\displaystyle \displaystyle \lambda^{'}(z) = -\gamma - \sum_{n=1}^{\infty} \{\frac{1}{n}\ (1+\frac{z}{n})^{-1} - \frac{1}{n} \}$ (6)
... so that is...
$\displaystyle \displaystyle \lambda^{'}(0) = -\gamma$ (7)
Deriving both terms of (6) we obtain...
$\displaystyle \displaystyle \lambda^{''}(z) = \sum_{n=1}^{\infty} \frac{1}{n^{2}}\ (1+\frac{z}{n})^{-2}$ (8)
... so that is...
$\displaystyle \displaystyle \lambda^{''}(0) = \zeta (2)$ (9)
... where $\displaystyle \displaystyle \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}}$ is the so called 'Riemann zeta function'. Proceeding along the way we find for $\displaystyle n>1$ ...
$\displaystyle \displaystyle \lambda^{(n)}(0) = (-1)^{n}\ (n-1)!\ \zeta(n)$ (10)
... so that the requested McLaurin expansion is...
$\displaystyle \displaystyle \ln z! = -\gamma\ z + \sum_{n=2}^{\infty} (-1)^{n}\ \zeta(n)\ \frac{z^{n}}{n}$ (11)
The (11) converges for $\displaystyle |z|<1$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$